Yes, there can be sub-intervals of length "n" with no prime numbers. however, you must prove this for all n.
And the second part is less straightforward, but still doable. The upper bound is always pi(n), because that is the maximum number of prime numbers in any sub-interval of the natural numbers of length n
?
In my previous post, I was addressing the second part of your question, not the first (I suppose I should've mentioned this). Basically, in your question you claim that if we have a subinterval of length n (has n elements), then there must exist a subinterval containing k primes for all k which exceeds 0 but is bounded by pi(x). In my previous post, I provided a counter example to this claim, showing how there exists a subinterval of length 3, but within the subinterval which I specifically mentioned (20,21,22) there is no subinterval which contains any primes. According to your question, within the subinterval (20,21,22), there should exist a subinterval which contains 1 prime and also another subinterval which contains 2 primes. Obviously, that's not true in this example.
Also, the first part of your question doesn't quite make sense either. For instance, the elements 2 and 3 form a subinterval of length 2, and clearly any subinterval within this subinterval must contain a prime. Thus your claim is false.
So this is why I'm asking for a clarification, as the question in its current wording isn't true.
. I actually forgot a term on the LHS from the diophantine equation I was trying to remember though lol, try the edited problem too.
, (2,1), (5,2), (13,5), (34,13), (89,34), \cdots $ and by symmetry, $(1,1), (1,2), (2,5), (5,13), (13,34), (34,89), \cdots $. The recurrence relation that generates these solutions define the odd-indexed Fibonacci numbers with those exact initial numbers, so the general solutions to the equation are $(1,1),(F_{2n-1},F_{2n+1}),(F_{2n+1},F_{2n-1}) \; \; \forall n \in \mathbb{N})
-f(i)| \leq 1 \; \forall i \in [a,b] \\ $If $f(a) <0< f(b), \exists c \in (a,b) : f(c) = 0)
=\{x+1,x+2,\ldots,x+n\}$ is given by $g_n(x):=\pi(x+n)-\pi(x)$ for any non-negative integer x.\\ \\ If we look at the shifted interval $I_n(x+1)$ instead, we might lose a prime from no longer having $x+1$ in our set, and we might gain a prime from adding $x+n+1$ to our set. \\ \\It might also be the case that both or neither of these things happen. In any case, the main point is that $g_n(x+1)-g_n(x)$ can either be $0$,$-1$ or $1$. Ie the function $g_n(x)$ takes steps of size at most $1$. \\ \\ By the discrete intermediate value theorem, it then suffices to show that for each $n$, we can find an $x_n$ with $g_n(x_n)=0$ (as we trivially have $g(0)=\pi(n)$, so every integer between $0$ and $\pi(n)$ must then lie in the range of $g$.)\\ \\ Proving this fact also establishes the first assertion, so we can now kill these two birds with the one stone next post.$)
 to this. Since the quadratic residue mod 3 is 0 or 1, the equation can only hold if a and b are both divisible by 3 (otherwise LHS would be 1 or 2 mod 3 whilst the RHS is 0 mod 3). So, a=3k, b=3j, c=3l,$ 9k^2+9j^2=3(9l^2). $Dividing through by 9, we find that (k,j,l) is a smaller natural solution. By Fermat's method of descent, there are no natural solutions. $ \\ x^3+2y^3=4z^3. $ Similar to first q, 2$|$x, and thus 2$|$y as 4$|$RHS and $x^3$ is even. But since the LHS is then divisible by 8, so is the RHS, so $2|z$. $(\frac{x}{2},\frac{y}{2}, \frac{z}{2})$ is a smaller natural solution, and thus there are no natural solutions. $\\$ Adding both equations,$ 7(a^2+b^2)=p^2+q^2. $Since the quadratic residue of 7 is 0,1,2,4, and the LHS is 0 mod7, 7 must divide both p and q. This implies that $7|a,b$ so (a/7,b/7,p/7,q/7) is a smaller natural solution, so no natural solutions exist. )
>2^5*k$. This shows that each term, is strictly at least twice as large as the previous term. $u_5$ is the first term to contain a fourth power of two, and hence, this sequence diverges into infinity.$)